Question

it has 2 digits.
the digit in the tens place is 2more than twice the digit in ones place.
the number formed by reversing the order is 5 more than three times the sum of the digits.
find the number

Answer 1

Let ten's digit be "M" and one's digit be "N".

The digit at ten's is 2 more than twice the digit in once.

According to question,

=> M = 2 + 2N ____ (eq 1)

The number formed by reversing the order is five more than three times the sum of digits.

Original number = 10M + N

Revered number = 10N + M

According to question,

=> 10N + M = 5 + 3(M + N)

=> 10N + M = 5 + 3M + 3N

=> 10N - 3N + M - 3M = 5

=> 7N - 2M = 5

=> 7N - 2(2 + 2N) = 5 [From (eq 1)]

=> 7N - 4 - 4N = 5

=> 3N = 5 + 3

=> 3N = 9

=> N = 3

Put value of N in (eq 1)

=> M = 2 + 2(3)

=> M = 2 + 6

=> M = 8

So,

Original number = 10M + N

=> 10(8) + 3

=> 83

∴ Original number is 83.

Answer 2

$suppose \: ten \: digit \: be \: x \: and \: y.$

$the \: digit \: in \: tens \: place \: is \: 2more \: than \: twice \: the \: digit \: in \: ones \: places.$

$x \: + 2 + 2y \: ......(eq1)$

$the \: number \: formed \: by \: reserving \: the \: order \: is \: 5more \: than \: three \: times \: the \: sum \: of \: the \: digits.$

$original \: no. = 10x \: + y$

$revered \: no. \: = 10y \: + \: x$

$= 10y \: + x \: = 5 + 3(x \: + \: y)$

$= 10y \: + x \: = 5 + 3x + 3y$

$= 10y \: - \: 3 \: + x \: - 3x \: = 5.$

$= 7y \: - \: 2x \: = 5.$

$= 7y \: - 2(2 + 2y) = 5(eq1).$

$= 7y \: - \: 4 \: - \: 4y \: = 5.$

$= 3y \: = 5 + 3.$

$= 3y \: = 9.$

$y \: = \: 3.$

$value \: of \: y \: in \: (eq1).$

$x \: = 2 + 2(3).$

$x \: = 2 + 6.$

$x \: = 8.$

$original \: no. \: = 10x \: + \: y \: .$

$= 10(8) \: + 3.$

$= 83.$